Let $X$ be a $n\times n$ matrix with $n\in\mathbb{N}^{+}$ and all off diagonal elements of $X$ following a uniform distribution $x_{ij}\sim\mathcal{U}(0,1)$ for any $i,j\in\{1,...,n\}$ with $i\neq j$. The diagonal elements of $X$ are set such that the rowsum is $0$ or in other words $$ x_{ii}=-\sum_{j=1,j\neq i}x_{ij} $$ I am checking if such matrices are almost surely diagonizable in general.
If so, my attempt would be to define the general form of the characteristic polynomial for any eigenvalue $\lambda_{i}$ with $i\in\{1,...,n\}$ $$ p_{X}(\lambda_{i})=\sum_{j=0}^{n}\lambda_{i}^{j}c_{j} $$ for some coefficients $c_{0},...,c_{n}\in\mathbb{R}$. The characteristic polynomial $p_{X}(\lambda_{i})$ has a root with multiplicity $2$ if $$ p_{X}(\lambda_{i})=(\lambda_{i}-a)^{2}\bar{p}_{X}(\lambda_{i})=(\lambda^{2}-2a\lambda+a^{2})\bar{p}_{X}(\lambda_{i}) $$ for some root $a\in\mathbb{R}$ with $$ \bar{p}_{X}(\lambda_{i})=\sum_{j=0}^{n-2}\lambda_{i}^{j}\bar{c}_{j} $$ again a polynomial. For this to hold in general for any $\lambda_{i}$ the following $3$ equations must be satisfied $$ \lambda^{n}_{i}=\lambda^{2}_{i}\lambda^{n-2}_{i}\bar{c}_{n-2} $$ $$ \lambda^{n-1}_{i}c_{n-1}=-2a\lambda_{i}\lambda^{n-2}_{i}\bar{c}_{n-2} $$ and $$ \lambda^{n-2}_{i}c_{n-2}=a^{2}\lambda^{n-2}_{i}\bar{c}_{n-2} $$ as $c_{n}=1$ and $c_{n-1}=\text{trace}(X)$ implying $\bar{c}_{n-2}=1$ and $\text{trace}(X)=-2$, therefore the probability of $X$ not being diagonizable boils down to the probability of the trace of $X$ being precisely equal to some $-2a=b\in\mathbb{R}$.
For any fixed $x_{ii}$ and $x_{1j}$ with $i,j=2,3,...,n$ the probability of $x_{11}$ being precisely $b-\sum_{i=2}^{n}x_{ii}$ is $0$ for continuous distributed random variables (you could go back to a Lebesgue integral over an atom is equal to $0$ if I recall correctly).
The probability of a root with multiplicity $2$ equal to $0$ implies the same for higher multiplicities as the polynomial could always be written in the form stated above displaying a root with multiplicity of $2$ multiplied by another polynomial $\bar{p}_{X}(\lambda_{i})$.
If my statement is true, does this proof make any sense?